Question

solve the given equation by variation of parameters.
y - 2y + y = e^x
the solution is y = \boxed{}.

Explanation:

Step1: Find the complementary function

The characteristic equation is \(r^{2}-2r + 1=0\), which factors as \((r - 1)^{2}=0\). So \(r=1\) (repeated root). The complementary function \(y_{c}=(C_{1}+C_{2}x)e^{x}\), and \(y_{1}=e^{x}\), \(y_{2}=xe^{x}\).

Step2: Calculate the Wronskian

\(W=

$$\begin{vmatrix}e^{x}&xe^{x}\\e^{x}&(x + 1)e^{x}\end{vmatrix}$$

=e^{x}\cdot(x + 1)e^{x}-xe^{x}\cdot e^{x}=e^{2x}\)

Step3: Find \(u_{1}\) and \(u_{2}\)

\(u_{1}=-\int\frac{y_{2}f(x)}{W}dx=-\int\frac{xe^{x}\cdot e^{x}}{e^{2x}}dx=-\int xdx=-\frac{x^{2}}{2}\)
\(u_{2}=\int\frac{y_{1}f(x)}{W}dx=\int\frac{e^{x}\cdot e^{x}}{e^{2x}}dx=\int 1dx=x\)

Step4: Construct the particular solution

\(y_{p}=u_{1}y_{1}+u_{2}y_{2}=-\frac{x^{2}}{2}e^{x}+x\cdot xe^{x}=\frac{x^{2}}{2}e^{x}\)

Step5: Write the general solution

\(y=y_{c}+y_{p}=(C_{1}+C_{2}x)e^{x}+\frac{x^{2}}{2}e^{x}\)

Answer:

\(y=(C_{1}+C_{2}x)e^{x}+\frac{1}{2}x^{2}e^{x}\)